Fundamental Concepts ICSE Class-7 Concise Selina Solutions

Fundamental Concepts ICSE Class-7 Concise Selina mathematics Solutions Chapter-11. We provide step by step Solutions of Exercise / lesson-11 Fundamental Concepts ( Including Fundamental Operations ) for ICSE Class-7 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-11 A , Exe-11 B Exe-11 C , Exe-11 D , Exe-11 E and    Exe-11 F to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7.

Fundamental Concepts ICSE Class-7 Concise Selina mathematics Solutions Chapter-11

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–: Select Topics :–

 

Exe-11 A,

Exe-11 B,

Exe-11 C,

Exe-11 D,

Exe-11 E,

Exe-11 F,

 

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Exercise – 11 A Fundamental Concepts ICSE Class-7 mathematics 

Question 1.

Separate constant terms and variable terms from tile following :

Answer

Question 2.

Constant is only 8 others are variables
(i) 2x ÷ 15
(ii) ax+ 9
(iii) 3x² × 5x
(iv) 5 + 2a-3b
(v) 2y –  z÷x
(vi) 3p x q ÷ z
(vii) 12z ÷ 5x + 4
(viii) 12 – 5z – 4
(ix) a³ – 3ab² x c
Answer

(i)

2x ÷ 15 = 2x/15

It is a monomial as it has one term.

(ii)

ax + 9: It is binomial

(∵ It has two terms)

(iii)

3x² × 5x = 15x³ – It is monomial

(∵ It has one term)

(iv)

5 + 2a – 3b – It is trinomial

(∵ It has three terms)

(v)

2y –(7/3) z ÷ x = 2y – (7z/3x) – It is binomial

(∵ It has two terms)

(vi)

3p x q ÷ z – (3pq/z) – It is monomial

(∵ It has one terms)

(vii)

12z ÷ 5x + 4 = (12z/5x)+4 – It is binomial

(∵ It has two terms)

(viii)

12 – 5 z – 4 = 8 – 5 z – It is binomial

(∵ It has two terms)

(ix)

a³ – 3ab² x c  = a³ – 3ab²c  – It is binomial

(∵ It has two terms)

Question 3.

Write the coefficient of:
(i) xy in – 3axy
(ii) z² in p²yz²
(iii) mn in -mn
(iv) 15 in – 15p²
Answer
(i) Co-efficient of xy in – 3 axy = – 3a

(ii) Co-efficient of z² in p²yz² = p²y

(iii) Co-efficient of mn in – mn = – 1

(iv) Co-efficient of 15 in – 15p² is -p²

Question 4.

For each of the following monomials, write its degree :
(i) 7y
(ii) – x²y
(iii) xy²z
(iv) – 9y²z³
(v) 3 m³n⁴
(vi) – 2p²q³r⁴
Answer

(i) Degree of 7y = 1

(ii) Degree of – x²y = 2 + 1 = 3 

(iii) Degree of xy²z = 1 + 2 + 1 = 4 

(iv) Degree of – 9y²z³ = 2 + 3 = 5 

(v) Degree of 3m³n⁴ = 3 + 4 = 7 

(vi) Degree of – 2p²q³r⁴ = 2 + 3 + 4 = 9 

Question 5.

Write the degree of each of the following polynomials :
(i) 3y³-x²y² + 4x
(ii) p³q² – 6p²q⁵ + p⁴q⁴
(iii) – 8mn⁶+ 5m³n
(iv) 7 – 3x²y + y²
(v) 3x – 15
(vi) 2y²z + 9yz³
Answer
(i) The degree of 3y³ – x²y²+ 4x is 4 as x² 

(ii) The degree of p³q² – 6p²q⁵ – p⁴q⁴ is 8 as p⁴ q⁴ is the term which has highest degree. 

(iii) The degree of – 8mn⁶ + 5m³n is 7 as – 8mx⁶ is the term that has the highest degree.

(iv) The degree of 7 – 3x²y + y² is 3 as – 3x²y is the term which has the highest degree. 

(v) The degree of 3x – 15 is 1 as 3x is the term which is highest degree.

(vi) The degree of 2y²z + 9yz³ is 4 as 9yz³ has the highest degree. 

Question 6.

Group the like term together :
(i) 9x², xy, – 3x², x² and – 2xy
(ii) ab, – a²b, – 3ab, 5a²b and – 8a²b
(iii) 7p, 8pq, – 5pq – 2p and 3p
Answer

(i) 9x², – 3x² and x² are like terms xy and – 2xy are like terms.

(ii)

ab, – 3ab, are like terms,

– a²b, 5a²b, – 8a²b are like terms

(iii)

7p, – 2p and 3p are like terms,

8pq, – 5pq are like terms.

Question 7.

Write numerical co-efficient of each of the followings :
(i) y
(ii) -y
(iii) 2x²y
(iv) – 8xy³
(v) 3py²
(vi) – 9a²b³
Answer

(i) Co-efficient of y = 1

(ii) Co-efficient of-y = – 1

(iii) Co-efficient of 2x²y is = 2 

(iv) Co-efficient of – 8xy³ is = – 8 

(v) Co-efficient of 3py² is = 3 

(vi) Co-efficient of – 9a²b³ is = – 9 

Question 8.

In -5x³y²z⁴; write the coefficient of:
(i) z²
(ii) y²
(iii) yz²
(iv) x³y
(v) -xy²
(vi) -5xy²z
Also, write the degree of the given algebraic expression.
Answer

(i)

-5x³y²z⁴

Co-efficient of z² is -5x³y²z²

Degree of the given expression is 3 + 2 + 4 = 9

(ii)

-5x³y²z⁴

Co-efficient of y² is -5x³z⁴

Degree of the given expression is 3 + 2 + 4 = 9

(iii)

-5x³y²z⁴

Co-efficient of yz² is -5x³yz²

Degree of the given expression is 3 + 2 + 4 = 9

(iv)

-5x³y²z⁴

Co-efficient of x³y is -5yz⁴

Degree of the given expression is 3 + 2 + 4 = 9

(v)

-5x³y²z⁴

Co-efficient of -xy² is 5x²z⁴

Degree of the given expression is 3 + 2 + 4 = 9

(vi)

-5x³y²z⁴

Co-efficient of -5xy²z is x²z³

Degree of the given expression is 3 + 2 + 4 = 9

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Fundamental Concepts ICSE Class-7 Exe-11 B

Question 1.

Fill in the blanks :
(i) 8x + 5x = ………
(ii) 8x – 5x =……..
(iii) 6xy² + 9xy² =……..
(iv) 6xy² – 9xy² = ………
(v) The sum of 8a, 6a and 5b = ……..
(vi) The addition of 5, 7xy, 6 and 3xy = …………
(vii) 4a + 3b – 7a + 4b = ……….
(viii) – 15x + 13x + 8 = ………
(ix) 6x²y + 13xy² – 4x²y + 2xy² = ……..
(x) 16x² – 9x² = and 25xy² – 17xy²=………

Answer

(i) 8x + 5x = 13x 

(ii) 8x – 5x = 3x 

(iii) 6xy² + 9xy² = 15xy² 

(iv) 6xy² – 9xy² = -3xy² 

(v) 

The sum of 8a, 6a and 5b

= 8a + 6a + 5b = 14a + 5b

(vi)

The addition of 5, 7xy, 6 and 3xy

= 5 + 6 + 7xy + 3xy = 11 + 10 xy

(vii)

4a + 3b – 7a + 4b

= 4a – 7a + 3b + 4b = -3a + 7b = 7b – 3a

(viii) – 15x + 13x + 8 = – 2x + 8 = 8 – 2x

(ix)

6x²y + 13xy² – 4x²y + 2xy²

= 6x²y – 4x²y + 13xy² + 2xy² = 2x²y + 15xy²

(x) 16x² – 9x² =  7x² and 25xy² – 17xy² = 8xy² 

Question 2.

Add :
(i)- 9x, 3x and 4x
(ii) 23y², 8y² and – 12y²
(iii) 18pq – 15pq and 3pq

Answer

(i)

– 9x, 3x and 4x

= – 9x + 7x = – 2x

(ii)

23y² , 8y² and – 12y²

= 31y² – 12y² = 19y²

(iii)

18pq – 15pq and 3pq

= 18pq + 3pq – 15pq = 21pq – 15pq = 6 pq

Question 3.

Simplify :
(i) 3m + 12m – 5m
(ii) 7n² – 9n² + 3n²
(iii) 25zy—8zy—6zy
(iv) -5ax² + 7ax² – 12ax²
(v) – 16am + 4mx + 4am – 15mx + 5am
Answer

(i)

3m + 12m – 5m

= 15m – 5m = 10m

(ii)

7n² – 9n² + 3n²

= 7n² + 3n² – 9n²

= 10n² – 9n² = n²

(iii)

25zy – 8zy  – 6zy

= 25zy – 14zy = 11zy

(iv)

-5ax² + 7ax² – 12ax²

= – 5ax² – 12ax² + 7ax²

= – 17ax² + 7ax² = – 10ax² 

(v)

– 16am + 4mx + 4am – 15mx + 5am

= – 16am + 4am + 5am + 4mx – 15mx

= – 16am + 9am + 4mx – 15mx

= – 7 am – 11 mx

Question 4.

Add :
(i) a + i and 2a + 3b
(ii) 2x + y and 3x – 4y
(iii)- 3a + 2b and 3a + b
(iv) 4 + x, 5 – 2x and 6x
Answer

(i)

a + b and 2a + 3b

=  a + b + 2a + 3b

= a + 2a + b + 3b = 3a + 4b

(ii)

2x + y and 3x – 4y

= 2x + 3x + y – 4y

= 5x – 3y

(iii)

– 3a + 2b and 3a + b

= – 3a + 3a + 2b + b

= 0 + 3b = 3b

(iv)

4 + x, 5 – 2x and 6x

= x – 2x + 6x + 4 + 5

= 7x – 2x + 9

= 5x + 9

Question 5.

Find the sum of:
(i) 3x + 8y + 7z, 6y + 4z- 2x and 3y – 4x + 6z
(ii) 3a + 5b + 2c, 2a + 3b-c and a + b + c.
(iii) 4x²+ 8xy – 2y² and 8xy – 5y² + x²
(iv) 9x² – 6x + 7, 5 – 4x and 6 – 3x²
(v) 5x² – 2xy + 3y² and – 2x² + 5xy + 9y²
and 3x² -xy- 4y²
(vi) a² + b² + 2ab, 2b² + c² + 2bc
and 4c²-a² + 2ac
(vii) 9ax – 6bx + 8, 4ax + 8bx – 7
and – 6ax – 46x – 3
(viii) abc + 2 ba + 3 ac, 4ca – 4ab + 2 bca
and 2ab – 3abc – 6ac
(ix) 4a² + 5b² – 6ab, 3ab, 6a² – 2b²
and 4b² – 5 ab
(x) x² + x – 2, 2x – 3x² + 5 and 2x² – 5x + 7
(xi) 4x³ + 2x² – x + 1, 2x³ – 5x²– 3x + 6, x² + 8 and 5x³ – 7x
Answer

(i)

3x + 8y + 7z; 6y + 4z- 2x and 3y – 4x + 6z

3x + 8y + 7z + 6y + 4z – 2x + 3y – 4x + 6z

= 3x – 6x + 17y + 17z

= – 3x + 17y + 17z

(ii)

3a + 5b + 2c, 2a + 3b-c and a + b + c.

3a + 5b + 2c + 2a + 3b – c + a + b + c.

= 6a + 9b + 3c – c

= 6a + 9b + 2c

(iii)

4x²+ 8xy – 2y² and 8xy – 5y² + x²

4x²+ 8xy – 2y² + 8xy – 5y² + x²

= 4x² + x² + 8xy + 8xy – 2y² – 5y²

= 5x² + 16xy – 7y²

(iv)

9x² – 6x + 7, 5 – 4x and 6 – 3x²

9x² – 6x + 7 + 5 – 4x + 6 – 3x²

= 9x² – 3x² – 6x – 4x + 7 + 5 + 6

= 6x² – 10x + 18

(v)

5x² – 2xy + 3y² and – 2x² + 5xy + 9y² and 3x² -xy- 4y²

5x² – 2xy + 3y² + – 2x² + 5xy + 9y² + 3x² -xy- 4y²

= 5x² – 2x² + 3x² – 2xy + 5xy – xy + 3y² + 9y² – 4y²

= 8x² – 2x² + 5xy – 3xy + 12y² – 4y²

=6x² + 2xy + 8y²

(vi)

a² + b² + 2ab, 2b² + c² + 2bc and 4c² -a² + 2ac

a² + b² + 2ab + 2b² + c² + 2bc + 4c² -a² + 2ac

= a² – a² + b² + 2b² + c² + 4c² + 2ab + 2bc + 2ac

= 3b² + 5c² + 2ab + 2bc + 2ac

(vii)

9ax – 6bx + 8, 4ax + 8bx – 7 and – 6ax – 46x – 3

9ax – 6bx + 8 + 4ax + 8bx – 7  – 6ax – 46x – 3

= 9ax + 4ax – 6ax – 6bx + 8bx – 4bx + 8 – 7 – 3

= 13ax – 6ax + 8bx – 10bx + 8 – 10

= 7ax – 2bx – 2

(viii)

abc + 2 ba + 3 ac, 4ca – 4ab + 2 bca and 2ab – 3abc – 6ac

abc + 2 ba + 3 ac + 4ca – 4ab + 2 bca + 2ab – 3abc – 6ac

= abc + 2abc – 3abc + 2ab – 4ab + 2ab + 3ca + 4ca – 6ca

= 3abc – 3abc + 4ab – 4ab + 7ca – 6ca

= 0 + 0 + ca = ca

(ix)

4a² + 5b² – 6ab, 3ab, 6a² – 2b² and 4b² – 5 ab

4a² + 5b² – 6ab + 3ab, 6a² – 2b² + 4b² – 5 ab

= 4a² + 6a² + 5b² – 2b² + 4b² – 6ab + 3ab – 5ab

= 10a² + 9b² – 2b² – 11ab + 3ab

= 10a² + 7b² – 8ab

(x)

x² + x – 2, 2x – 3x² + 5 and 2x² – 5x + 7

= x² + x – 2 + 2x – 3x² + 5 + 2x² – 5x + 7

= x² – 3x² + 2x² + x + 2x – 5x – 2 + 5 + 7

= 3x² – 3x² + 3x – 5x – 2 + 12

= 0 – 2x + 10

= – 2x + 10

(xi)

4x³ + 2x² – x + 1, 2x³ – 5x² – 3x + 6, x2 + 8 and 5x³ – 7x

4x³ + 2x² – x + 1 + 2x³ – 5x² – 3x + 6 + x² + 8 + 5x³ – 7x

= 4x³ + 2x³ + 5x³ + 2x² – 5x² + x² – x – 3x – 7x + 1 + 6 + 8

= 11x³ + 3x² – 5x² – 11x + 15

= 11x³ – 2x² – 11x + 15

Question 6.

Find the sum of:
(i) x and 3y
(ii) -2a and +5
(iii) – 4x² and +7x
(iv) +4a and -7b
(v) x³+3x²y and 2y²
(vi) 11 and -by
Answer

(i) x + 3y

(ii) -2a + 5

(iii) – 4x² + 7x

(iv) + 4a and -7b = 4a – 7b

(v) x³+3x²y and 2y² = x³+ 3x²y + 2y² 

(vi) 11 and -by = 11 – by

Question 7.

The sides of a triangle are 2x + 3y, x + 5y and 7x – 2y, find its perimeter.
Answer

Sides of a triangle are 2x + 3y, x + 5y, 7x – 2y

∴ Perimeter = sum of three sides of the triangle

= 2x + 3y + x + 5y + 7x – 2y

= 2x + x + 7x + 3y + 5y – 2y

= 10x + 8y – 2x

= 10x + 6y

Question 8.

The two adjacent sides of a rectangle are 6a + 96 and 8a – 46. Find its, perimeter.
Answer

Sides of a rectangle are 6a + 9b

and 8a – 4b

Let, length = 6a + 9b

and breadth = 8a – 4b

∴ Perimeter = 2 (l + b)

= 2 (6a + 9b + 8a – 4b)

= 2 (14 a + 5 b)

= 28a + 10b

Question 9.

Subtract the second expression from the first:

(i) 2a + b, a + b

(ii) – 2b + 2c, b + 3c

(iii) 5a + b, – 6b + 2a

(iv) a³ – 1 + a, 3a – 2a² 

(v) p + 2, 1

(vi) x + 2y + z, – x – y – 3z

(vii) 3a² – 8ab – 2b² , 3a² – 4ab + 6b²

(viii) 4pq – 6p² – 2q² , 9p²

(ix) 10abc, 2a² + 2abc – 4b²

(x) a² + ab + c², a² – d²

Answer

(i)

(2a + b) – (a + b)

= 2a + b – a – b

= 2a – a + b – b

= a + 0 = a

(ii)

(- 2b + 2c) + (b + 3c)

= – 2b + 2c – b – 3c

= – 2b – b + 2c – 3c

= – 3b – c

(iii)

(5a + b) – (- 6b + 2a)

= 5a + b + 6b – 2a

= 5a – 2a + b + 6b

= 3a + 7b

(iv)

(v) (p + 2) – 1 = p + 2 – 1 = p + 1

(vi)

(x + 2y + z) – (- x – y – 3z)

= x + 2y + z + x + y+ 3z

= x + x + 2y + y + z + 3z

= 2x + 3y + 4z

(vii)

(3a² – 8ab – 2b²) – (3a² – 4ab + 6b² )

= 3a² – 8ab – 2b² – 3a² + 4ab – 6b²

= 3a² – 3a² – 2b² – 6b² – 8ab + 4ab

= 0 – 8b² – 4ab

= – 4ab – 8b²

(viii)

(4pq – 6p² – 2q²) – (9p²)

= 4pq – 6p² – 2q² – 9p² 

= 4pq – 15p² – 2q²

(ix)

10abc – (2a² + 2abc – 4b² )

= 10 abc – 2a² – 2abc + 4b²

= 10 abc – 2 abc – 2a² + 4b²

= 8abc – 2a² + 4b²

(x)

(a² + ab + c²) – (a² – d² )

= a² + ab + c² – a² + d²

= a² – a² + ab + c² + d²

= ab + c² + d²

Question 10.

Subtract:

(i) 4x from 8 – x

(ii) – 8c from c + 3d

(iii) – 5a – 2b from b + 6c

(iv) 4p + p² from 3p² – 8p

(v) 5a – 3b + 2c from 4a – b – 2c

(vi) – xy + yz – zx from xy – yz – xz

(vii)  2x² – 7xy – y² from 3x² – 5xy + 3y²

(viii)  a² – 3ab – 6b² from 2b² – a² + 2ab

(ix) 4x² – 5x²y + y² from – 3y² + 5xy² – 7x² – 9x²y

(x) 6m³ + 4m² + 7m – 3 from 3m³ + 4

Answer

(i)

4x from 8 – x

= (8 – x) – 4x

= 8 – x – 4x

= 8 – 5x

(ii)

– 8c from c + 3d

= (c + 3d) – (- 8c)

= c + 3d + 8c

= 9c + 3d

(iii)

– 5a – 2b from b + 6c

= (b + 6c) – (- 5a – 2b)

= b + 6c + 5a + 2b

= 5a + 3b + 6c

(iv)

4p + p² from 3p² – 8p

= (3p² – 8p) – (4p + p²)

= 3p² – 8p – 4p – p²

= 2p² – 12p

(v)

5a – 3b + 2c from 4a – b – 2c

= (4a – b – 2c) – (5a – 3b + 2c)

= 4a – b – 2c – 5a + 3b – 2c

= 4a – 5a – b + 3b – 2c – 2c

= – a + 2b – 4c

(vi)

– xy + yz – zx from xy – yz – xz

= (xy – yz – xz) – (- xy + yz – xz)

= xy – yz + zx + xy – yz + xz

= xy + xy – yz – yz + zx + xz

= xy + xy – yz – yz + zx + xz

= 2(xy – yz + zx)

(vii)

2x² – 7xy – y² from 3x² – 5xy + 3y²

= (3x² – 5xy + 3y²) – (2x² – 7xy – y²)

= 3x² – 5xy + 3y² – 2x² + 7xy + y²

= 3x² – 2x² – 5xy + 7xy + 3y² + y²

= x² + 2xy + 4y²

(viii)

(ix)

4x² – 5x²y + y² from – 3y² + 5xy² – 7x² – 9x²y

= (- 3y² + 5xy² – 7x² – 9x²y) – (4x² – 5x²y + y²)

= – 3y² + 5xy² – 7x² – 9x²y – 4x² + 5x²y  – y²

= – 3y² – y² + 5xy² – 7×2 – 4x² – 9x²y + 5x²y

= – 4y² + 5xy² – 11x² – 4x²y

(x)

6m³ + 4m² + 7m – 3 from 3m³ + 4

= (3m³ + 4) – (6m³ + 4m² + 7m – 3)

= 3m³ + 4 – 6m³ + 4m² + 7m – 3

= 3m³ – 6m3 – 4m² – 7m + 4 + 3

= – 3m³ – 4m² – 7m + 7

Question 11.

Subtract – 5a² – 3a + 1 from the sum of 4a² + 3 – 8a and 9a – 7.
Answer

sum of 4a² + 3 – 8a and 9a – 7

= 4a² + 3 – 8a + 9a – 7 = 4a² + a – 4

∴ (4a² + a – 4) – (- 5a² – 3a + 1)

= 4a²  + a – 4 + 5a² + 3a – 1

= 4a² + 5a² + a + 3a – 4 – 1

= 9a² + 4a – 5

Question 12.

By how much does 8x³ – 6x² + 9x – 10 exceed 4x³ + 2x² + 7x -3 ?
Answer

8x³ – 6x² + 9x – 10 exceeds 4x³ + 2x² + 7x -3

= (8x³ – 6x² + 9x – 10) – (4x³ + 2x² + 7x – 3)

= 8x³ – 6x² + 9x – 10 – 4x³ + 2x² + 7x – 3

= 8x³ – 4x³ – 6x² – 2x² + 9x – 7x – 10 + 3

= 4x³ – 8x² + 2x – 7

Question 13.

What must be added to 2a³ + 5a – a² – 6 to get a² – a – a³ + 1 ?
Answer

We get, the required result by subtracting

2a³ + 5a – a² – 6 from – a³ + a² – a + 1

= (- a³ + a² – a + 1) – (2a³  – a² + 5a – 6)

= – a³ + a² – a + 1 – 2a³  + a² – 5a + 6

= – a³  – 2a³ + a² + a² – a – 5a + 1 + 6

= – 3a³ + 2a² – 6a + 7

Question 14.

What must be subtracted from a² + b² + lab to get – 4ab + 2b² ?
Answer

We get, the required result by subtracting

– 4ab + 2b² from a² + b² + 2ab

= a² + b² + 2ab – (- 4ab + 2b²)

= a² + b² + 2ab + 4ab – 2b²

= a² + b² – 2b² + 2ab + 4ab

= a² – b² + 6ab

Question 15.

Find the excess of 4m² + 4n² + 4p² over m²+ 3n² – 5p²
Answer

The required result will be by substracting

m²+ 3n² – 5p² from  4m² + 4n² + 4p²

= 4m² + 4n² + 4p² – (m² + 3n² – 5p²)

= 4m² + 4n² + 4p² – m² – 3n² – 5p²

= 4m² – m² + 4n² – 3n² + 4p² + 5p²

= 3m² + n² + 9p² 

Question 16.

By how much is 3x³ – 2x²y + xy² -y³ less than 4x³ – 3x²y – 7xy² +2y³
Answer

We can get the required results by substracting

3x³ – 2x²y + xy² -y³ from 4x³ – 3x²y – 7xy² +2y³

= (4x³ – 3x²y – 7xy² +2y³ ) – (3x³ – 2x²y + xy² -y³)

= 4x³ – 3x²y – 7xy² +2y³ – 3x³ + 2x²y – xy² + y³ 

= 4x³ – 3x³ – 3x²y + 2x²y – 7xy² – xy² +2y³ + y³ 

= x³ – x²y – 8xy² + 3y³

Question 17.

Subtract the sum of 3a² – 2a + 5 and a² – 5a – 7 from the sum of 5a² -9a + 3 and 2a – a² – 1
Answer

Sum of 3a² – 2a + 5 and a² – 5a – 7

= 3a² – 2a + 5 + a² – 5a – 7

= 3a² + a² – 2a – 5a + 5 – 7

= 4a² – 7a – 2

and sum of 5a² -9a + 3 and 2a – a² – 1

= 5a² – 9a + 3 + 2a – a² – 1

= 5a² – a² – 9a + 2a + 3 – 1

= 4a² – 7a + 2

Now (4a² – 7a + 2) – (4a² – 7a – 2)

= 4a² – 7a + 2 – 4a² + 7a + 2

= 4a² – 4a²  – 7a + 7a + 2 + 2

= 0 + 0 + 4 = 4

Question 18.

The perimeter of a rectangle is 28x³+ 16x² + 8x + 4. One of its sides is 8x² + 4x. Find the other side
Answer

Question 19.

The perimeter of a triangle is 14a² + 20a + 13. Two of its sides are 3a² + 5a + 1 and a² + 10a – 6. Find its third side.
Answer

perimeter of a triangle = 14a² + 20a + 13

Sum of two sides

= 3a² + 5a + 1 + a² + 10a – 6

= 3a² + a² + 5a + 10a + 1 – 6

= 4a² + 15a – 5

∴ Third side = (14a² + 20a + 13) – (4a² + 15a – 5)

= 14a² + 20a + 13 – 4a² – 15a + 5

= 14a² – 4a² + 20a – 15a + 13 + 5

= 10a² + 5a + 18

Question 20.

(i) If x = 4a² + b² – 6ab; y = 3b² – 2a² + 8ab and z = 6a² + 8b² – 6ab

find: x + y + z

(ii) If x = 4a² + b² – 6ab; y = 3b² – 2a² + 8ab and z = 6a² + 8b² – 6ab

find: x – y – z
Answer

(i)

x = 4a² + b² – 6ab
y = 3b² – 2a² + 8ab
z = 6a² + 8b² – 6ab

x + y + z

= 4a² + b² – 6ab + 3b² – 2a² + 8ab + 6a² + 8b² – 6ab

= 4a² – 2a² + 6a² + b² + 3b² + 8b² – 6ab + 8ab – 6ab

= 10a² – 2a² + 12b² – 12ab + 8ab

= 8a² + 12b² – 4ab

(ii)

x = 4a² + b² – 6ab
y = 3b² – 2a² + 8ab
z = 6a² + 8b² – 6ab

x – y – z

= (4a² + b² – 6ab) – (3b² – 2a² + 8ab) – (6a² + 8b² – 6ab)

= 4a² + b² – 6ab – 3b² + 2a² – 8ab – 6a² – 8b² + 6ab

= 4a² – 2a² – 6a² + b² – 3b² – 8b² – 6ab – 8ab + 6ab

= 6a² – 6a² + b² – 11b² – 14ab + 6ab

= – 10b² – 8ab

Question 21.

Answer

(i)

m = 9x² – 4xy + 5y²

n = – 3x² + 2xy – y²

∴ 2m – n

= 2 (9x² – 4xy + 5y²) – (- 3x² + 2xy – y²)

= 18x² – 8xy + 10y² + 3x² – 2xy + y²

= 18x² + 3x² – 8xy – 2xy + 10y² + y² 

= 21x² – 10xy + 11y²

(ii)

m = 9x² – 4xy + 5y²

n = – 3x² + 2xy – y²

∴ m + 2n

= (9x² – 4xy + 5y²) + 2 (- 3x² + 2xy – y²)

= 9x² – 4xy + 5y² – 6x² + 4xy – 2y²

= 9x² – 6x² – 4xy + 4xy + 5y² – 2y²

= 3x² + 3y²

(iii)

m = 9x² – 4xy + 5y²

n = – 3x² + 2xy – y²

∴ m – 3n

= (9x² – 4xy + 5y²) – 3 (- 3x² + 2xy – y²)

= 9x² – 4xy + 5y² + 9x² – 6xy + 3y²

= 18x² – 10xy + 8y²

Question 22.

Simplify:

Answer

(i)

3x + 5(2x + 6) – 7x

⇒ 3x + 10x + 30 – 7x

⇒ 3x + 10x – 7x + 30

⇒ 13x – 7x + 30

⇒ 6x + 30

(ii)

3(4y – 10)  2(y – 1)

⇒ 12y – 30 + 2y – 2

⇒ 12y + 2y – 30 – 2

⇒ 14y – 32

(iii)

– (7 + 6x) – 7(x + 2)

⇒ – 7 – 6x – 7x – 14

⇒ – 7x – 6x – 7 – 14

⇒  -13 x – 21

(iv)

x – (x – y) – y – (y – x)

⇒ x – x + y – y – y + x

⇒ 2x – x – 2y + y

⇒ x – y

(v)

4x + 7y – (5y – 8) – 2x

⇒ 4x + 7y – 5y + 8 – 2x

⇒ 4x – 2x + 7y – 5y + 8

⇒ 2x + 2y + 8

(vi)

– 2m + 5 + 4(m – 3)

⇒ – 2m + 5 + 4m – 12

⇒ – 2m + 4m + 5 – 12

⇒ 2m – 7

(vii)

2x – y + 5 – (x – y)

⇒ 2x – y + 5 – (x – y)

⇒ 2x – x + 5

⇒ x + 5

(viii)

2(x – y) – (x – 8)

⇒ 2x – 2y – x + 8

⇒ 2x – x – 2y + 8

⇒ x – 2y + 8

(ix)

4(3x – 8) – 3(5x + 3) – 2(6x – 8)

⇒ 12x – 32 – 15x – 9 – 12x + 16

⇒ 12x – 15x – 12x – 32 – 9 + 16

⇒ 12x – 27x – 41 + 16

⇒ – 15 x – 25

(x)

5(x – 4) – 3(x – 4) + 7(x – 4)

⇒ 5x – 20 – 3x + 12 + 7x – 28

⇒ 5x + 7x – 3x + 12 + 7x – 28

⇒ 12x – 3x – 48 + 12

⇒ 9x – 36

 

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Exe – 11 C Soved Questions of Fundamental Concepts for ICSE Class-7 

Question 1.

Multiply:

Answer

(i)

Product of 3x, 5x²y and 2y

= 3x + 5x²y × 2y

= 3 × 5 × 2 × x × x² × y × y

= 30x³y²

(ii)

Product of 5, 3a and 2ab²

= 5 × 3a × 2ab²

= 5 × 3 × 2 × a × ab²

= 30a²b²

(iii)

Product of 5x + 2y and 3xy

= 3xy (5x + 2y)

= 3xy × 5x + 3xy × 2y

= 15x²y + 6xy²

(iv)

Product of 6a – 5b and – 2a

= -2a (6a – 5b)

= – 2a × 6a + (- 2a) × (- 5b)

= – 12a² + 10ab

(v)

Product of 4a + 5b and 4a – 5b

= 16a² – 25b²

4a + 5b

(vi)

Product of 9xy + 2y² and 2x – 3y

= 18x²y – 23xy² – 6y³

9xy + 2y²

(vii)

Product of – 3m²n + 5mn – 4mn² and 6m²n

= 6m²n (- 3m²n + 5mn – 4mn² )

= 6m²n × (- 3m²n) + 6m²n × 5mn + 6m²n ×  (- 4mn²)

= 18m⁴n² + 30m³n² – 24 m³n³

(viii)

Product of 6xy² – 7x²y² + 10x³ and – 3x²y³

= – 3x²y³ (6xy² – 7x²y² + 10x³)

= – 3x²y³  × 6xy²  + (- 3x²y³) × (- 7x²y² ) + (- 3x²y³) × 10x³

= – 18x³y⁵ + 21x⁴y⁵ – 30x⁵y³

Question 2.

Copy and complete the following multi-plications :

Answer

(i) 

3a + 2b
× – 3xy
– 9axy – 6bxy  

(ii) 

9x + 5y
× – 3xy
– 27x²y + 15xy² 

(iii) 

3xy – 2x² – 6x
×     -5x²y
-15x³y² + 10x⁴y + 30x³y

(iv) 

a + b
× a + b
a² + ab
ab + b² 
a² + 2ab + b² 

(v) 

ax – b
× 2ax + 2b²
2a²x² – 2abx + 2ab²x – 2b³ 

(vi) 

2a – b + 3c
× 2a – 4b
4a2 – 2ab + 6ac
– 8ab + 4b² – 12bc
4a² – 10ab + 6ac + 4b² – 12bc

(vii) 

3m² + 6m – 2n
× 5n – 3m
15m²n + 30mn – 10n² – 9m³ – 18m²
+ 6mn
15m²n + 36mn – 10n² – 9m³ – 18m²

(viii)

(ix) 

4x³ – 10x² + 6x – 8
× 3 + 2x – x²
12x³ – 30x² + 18x – 24
8x⁴ – 20x³ + 12x² – 16x
4x⁵ + 10x⁴ – 6x³ + 8x² 
4x⁵ + 18x⁴ – 14x³ – 10x² + 2x – 24  

Question 3.

Evaluate :

Answer

(i)

(c + 5)(c – 3) = c (c – 3) +5 (c – 3)

= c² – 3c + 5c – 15

= c² + 2c – 15

(ii)

(3c – 5d)(4c – 6d)

= 3c (4c – 6d) – 5d(4c – 6d)

= 12c² – 18cd – 20cd + 30d²

= 12c² – 38cd + 30d²

(iii) 

(iv)

(a² + 2ab + b²)(a + b)

= a(a² + 2ab + b²) + b(a² + 2ab + b²)

= a³ + 2a²b + ab² + a²b + 2ab² + b³

= a³ + 3a²b + 3ab² + b³

(v)

(3x – 1)(4x³ – 2x² + 6x – 3)

= 3x (4x³ – 2x² + 6x – 3) – 1 (4x³ – 2x² + 6x – 3)

= 12x⁴ – 6x³ + 18x² – 9x – 4x³ + 2x² – 6x + 3

= 12x⁴ – 6x³ – 4x³ + 18x² + 2x² – 9x – 6x + 3

= 12x⁴ – 10x³ + 20x² – 15x + 3

(vi)

(4m – 2)(m² + 5m – 6)

= 4m (m² + 5m – 6) – 2(m² + 5m – 6)

= 4m³ + 20m² – 24m – 2m² – 10m + 12

= 4m³ + 20m² – 2m² – 24m – 10m + 12

= 4m³ + 18m² – 34m + 12

(vii)

(8 – 12x + 7x² – 6x³)(5 – 2x)

= 5(8 – 12x + 7x² – 6x³) – 2x (8 – 12x + 7x² – 6x³)

= 40 – 60x + 35x² – 30x³ – 16x + 24x⁴ – 14x³ + 12x⁴

= 40 – 60x – 16x + 35x² + 24x⁴ – 30x³ – 14x³ + 12x⁴

= 40 – 76x + 59x² – 44x³ + 12x⁴ 

(viii)

(4x² – 4x + 1)(2x³ – 3x² + 2)

= 4x² (2x³ – 3x² + 2) – 4x (2x³ – 3x² + 2) + 1 (2x³ – 3x² + 2)

= 8x⁵ – 12x⁴ + 8x² – 8x⁴ + 12x³ – 8x + 2x³ – 3x² + 2

= 8x⁵ – 12x⁴ – 8x⁴ + 12x³ + 2x³ + 8x² – 3x² – 8x  + 2

= 8x⁵ – 20x⁴ + 14x³ + 5x² – 8x + 2

(ix)

(6p² – 8pq + 2q²) (- 5p)

= -5p × 6p² – 5p × (- 8pq) – 5p (2q²)

= – 30p³ + 40p²q – 10pq²

(x)

– 4y (15 + 12y – 8z) (x – 2y)

= – 4y (x – 2y)(15x + 12y – 8z)

= (- 4xy + 8y²)(15x + 12y – 8z)

= – 4xy (15x + 12y – 8z) + 8y² (15x +12y – 8z)

= -60x²y – 48xy² + 32xy + 120xy² + 96y³ – 64y²z

= -60x²y – 48xy² + 120xy² – 64y²z + 96y³ + 32xy

= -60x²y + 72xy2 – 64y²z + 96y³ + 32xyz

(xi)

(a² + b² + c² – ab – bc – ca)(a + b + c)

= a (a² + b² + c² – ab – bc – ca) + b (a² + b² + c² – ab – bc – ca) + c (a² + b² + c² – ab – bc – ca)

= a³ + ab² + ac² – a²b – abc – ca² +a²b + b³ + bc² – ab² – b²c  – abc + a²c + b²c + c³ – abc – bc² – c²a

= a³ + b³ + c³ -a²b + a²b – ca² + a²c + bc² – bc² – ab² + ab² – abc – abc – abc + ac² – ac² + b²c – b²c

= a³ + b³ + c³ – 3abc

Question 4.

Evaluate:

Answer

(i)

(a + b)(a – b)

= a (a – b) + b(a – b)

= a² – ab + ab – b²

= a² – b²

(ii) 

(a² + b²)(a + b)(a – b)

= (a² + b²)(a² – b²)    …{from(i)}

= a2 (a² – b²) + b² (a² – b²)

= a⁴ – a²b² + a²b² – b⁴

= a⁸ – a⁴b⁴ + a⁴b⁴ – b⁸

= a⁴ – b⁴

(iii) 

(a⁴ + b⁴)(a² + b²)(a + b)(a – b)

= (a⁴ + b⁴) (a⁴ – b⁴)  ….{from(ii)}

= a⁴ (a⁴ + b⁴) + b⁴ (a⁴ + b⁴)

= a⁸ – a⁴b⁴ + a⁴b⁴ – b⁸

= a⁸ – b⁸

Question 5.

Evaluate :

Answer

(i)

(3x – 2y)(4x + 3y)

= 3x (4x + 3y) – 2y (4x + 3y)

= 12x² + 9xy – 8xy – 6y² 

= 12x² + xy – 6y²

(ii)

(3x – 2y)(4x + 3y)(8x – 5y)

= 3x (4x + 3y) – 2y (4x + 3y)(8x – 5y)

= (12x² + 9xy – 8xy – 6y² )(8x – 5y)

= (12x² + xy – 6y² )(8x – 5y)

= 8x (12x² + xy – 6y² ) – 5y (12x² + xy – 6y² )

= 96x³ + 8x²y – 48xy² – 60x²y – 5xy² + 30y³

= 96x³ + 8x²y – 60x²y – 48xy² – 5xy² + 30y³

= 96x³ – 52x²y – 53xy² + 30y³

(iii)

(a + 5)(3a – 2)(5a + 1)

= {a (3a – 2) + 5(3a – 2)} (5a + 1)

= (3a² – 2a + 15a – 10)(5a + 1)

= (3a² + 13a – 10)(5a + 1)

= 5a (3a² + 13a – 10) + 1 (3a² + 13a – 10)

= 15a³ + 65a² – 50a + 3a² + 13a – 10

= 15a³ + 68a² – 37a – 10

(iv)

(a + 1)(a² – a + 1) and (a – 1)(a² + a + 1)

= a (a² – a + 1) + 1 (a² – a + 1)

= a³ – a² + a + a² – a + 1

= a³ + 1

(a – 1)(a² + a + 1)

= a(a² + a + 1) – 1(a² + a + 1)

= a³ + a² + a – a² – a – 1

= a³ – 1

Now, (a + 1)(a² – a + 1) + (a – 1)(a² + a + 1)

= a³ + 1 + a³ – 1

= 2a³

(v)

(5m – 2n)(5m + 2n)(25m² + 4n²)

= {5m (5m + 2n) – 2n(5m + 2n)} (25m² + 4n²)

= (25m² + 10mn – 10mn – 4n²)(25m² + 4n²)

= (25m² – 4n²)(25m² + 4n²)

= 25m² (25m² + 4n²) – 4n² (25m² + 4n²)

= 625m⁴ + 100m²n² – 100m²n² – 16n⁴

= 625m⁴ – 16n⁴

Question 6.

Multiply:

Answer

(i)

mn⁴, m³n and 5m²n³

⇒ 5m²n³ × mn⁴ × m³n

⇒5m²⁺¹⁺³ n³⁺⁴⁺¹

= 5m⁶n⁸ 

(ii)

2mnpq, 4mnpq and 5mnpq

⇒ 5mnpq × 2mnpq  × 4mnpq

⇒5×2×4m¹⁺¹⁺¹  n¹⁺¹⁺¹  p¹⁺¹⁺¹ q¹⁺¹⁺¹

⇒ 40m³n³p³q³ 

(iii)

pq – pm and p²m

⇒ p²m × (pq – pm)

⇒ p³qm – p³m²

(iv)

x³ – 3y³ and 4x²y²

⇒ 4x²y² × (x³ – 3y³)

⇒ 4x⁵y² – 12x²y⁵

(v)

a³ – 4ab and 2a²b

⇒ 2a²b × (a³ – 4ab)

⇒ 2a⁵b – 8a³b²

(vi)

x² + 5yx – 3y² and 2x²y

⇒ 2x²y × (x² + 5yx – 3y²)

⇒ 2x⁴y + 10x³y² – 6x²y³

Question 7.

Multiply:

Answer

(i)

(2x + 3y)(2x + 3y)

⇒ 2x (2x + 3y) + 3y(2x + 3y)

⇒ 4x² + 6xy + 6xy + 9y² 

⇒ 4x² + 12xy + 9y²

(ii)

(2x – 3y)(2x + 3y)

⇒ 2x (2x + 3y) – 3y(2x + 3y)

⇒ 2x × 2x + 2x × 3y – 3y × 2x – 3y × 3y

⇒ 4x² + 6xy – 6xy – 9y²

⇒ 4x² + 0 – 9y²

⇒ 4x² – 9y²

(iii)

(2x – 3y)(2x + 3y)

⇒ 2x (2x + 3y) – 3y(2x + 3y)

⇒ 2x × 2x + 2x × 3y – 3y × 2x – 3y × 3y

⇒ 4x² + 6xy – 6xy – 9y²

⇒ 4x² + 0 – 9y²

⇒ 4x² – 9y²

(iv)

(2x – 3y)(2x – 3y)

⇒ 2x (2x – 3y) – 3y(2x – 3y)

⇒ 2x × 2x – 2x × 3y + 3y × 2x + 3y × 3y

⇒ 4x² – 6xy – 6xy + 9y²

⇒ 4x² – 12xy + 9y²

(v)

(- 2x + 3y)(2x – 3y)

⇒ – 2x (2x – 3y) + 3y(2x – 3y)

⇒ – 4x² + 6xy + 6xy – 9y²

⇒ – 4x² + 12xy – 9y²

(vi)

(vii)

(x – a)(x + 3b)

⇒ x (x + 3b) – a (x + 3b)

⇒ x² + 3bx – ax – 3ab

(viii)

(2x + 5y + 6)(3x + y – 8)

⇒ 2x (3x + y – 8) + 5y(3x + y – 8) + 6 (3x + y – 8)

⇒ 6x² + 2xy – 16x + 15xy + 5y² – 40y + 18x + 6y – 48

⇒ 6x² + 2xy + 15xy – 16x + 18x + 5y² – 40y + 6y – 48

⇒ 6x² + 17xy + 2x + 5y² – 34y – 48

(ix)

(3x – 5y + 2)(5x – 4y – 3)

⇒ 3x (5x – 4y – 3) – 5y (5x – 4y – 3) + 2 (5x – 4y – 3)

⇒ 15x² – 12xy – 9x – 25xy + 20y² + 15y + 10x – 8y – 6

⇒ 15x² – 12xy – 25xy – 9x + 10x + 20y² + 15y – 8y – 6

⇒ 15x² – 37xy + x + 20y² + 7y – 6

(x)

(6x – 2y)(3x – y)

⇒ 6x (3x – y) – 2y (3x – y)

⇒ 18x² – 6xy – 6xy + 2y²

⇒ 18x² – 12xy + 2y²

(xi)

(1 + 6x² – 4x³)(-1 + 3x – 3x²)

⇒ 1(- 1 + 3x – 3x²) + 6x² (- 1 + 3x – 3x²) – 4x³ (- 1 + 3x – 3x²)

⇒ – 1 + 3x – 3x² – 6x² + 18x³ – 18x⁴ + 4x³ – 12x⁴ + 12x⁵

⇒ – 1 + 3x – 9x² + 22x³ – 30x⁴ + 12x⁵

 

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 ICSE Class-7 Concise Selina mathematics Solutions Exe- 11 D 

Question 1.

Divide:

Answer

(i)

(ii)

(iii)

8x + 24 by 4

(iv)

4a² – a by – a

= (4a²-a)/-a

= (4a² /-a) – (a-a)

= – 4a + 1

(v)

8m – 16 by – 8

(vi) 

– 50 + 40p by 10p

(vii) 4x³ – 2x² by – x 

(viii)

10a³ – 15a²b by – 5a²

(ix) 12x³y – 8x²y² + 4x²y³ by 4xy 

= 3x² – 2xy + xy² 

(x) 9a⁴b – 15a³b² + 12a²b³ by – 3a²b 

= – 3a² + 5ab – 4b² 

Question 2.

Divide :

Answer

(i)

9a⁴b – 15a³b² + 12a²b³ by – 3a²b

= – 3a² + 5ab – 4b²

(ii)

m² – 2mn + n² by m – n

m – n

= m – n

(iii)

a² + 4a + 1 by 2a + 1

2a + 1

= 2a + 1

(iv)

p² + 4p + 4 by p + 2

p + 2

= p + 2

(v)

x² + 4xy + 4y² by x + 2y

x + 2y

= x + 2y

(vi)

2a² – 11a + 12 by a – 4

2a – 3

(vii)

6x² + 5x – 6 by 2x + 3

3x – 2

= 3x – 2

(viii)

8a² + 4a – 60 by 2a – 5

4a + 12

= 4a + 12

(ix)

9x² – 24xy + 16y² by 3x- 4y

3x – 4y

= 3x – 4y

(x)

15x² + 31xy + 14y² by 5x + 7y

3x + 2y

= 3x + 2y

(xi)

35a³ + 3a²b – 2ab² by 5a – b

7a² + 2ab

= 7a² + 2ab

(xii)

6x³ + 5x² – 21x + 10 by 3x – 2

2x² + 3x – 5

= 2x² + 3x – 5

Question 3.

The area of a rectangle is 6x²– 4xy – 10y² square unit and its length is 2x + 2y unit. Find its breadth
Answer

Area of a rectangle

= 6x² – 4xy – 10y² sq.units

Length = 2x + 2y units

= 3x – 5y units

Hence breadth = 3x – 5y units

Question 4.

The area of a rectangular field is 25x² + 20xy + 3y² square unit. If its length is 5x + 3y unit, find its breadth, Hence find its perimeter.
Answer

Area of a rectangle

= 25x² + 20xy + 3y²

Length = (5x + 3y) units

Hence Breadth = 5x + y

Hence perimeter of rectangular field

= 2 (l + b)

= 2 (5x + 3y + 5x + y)

= 2 (10x + 4y)

= 20x + 8y

Question 5.

Divide:

Answer

(i)

2m³n⁵ by – mn

= 2m³n⁵ / -mn

= – 2m²n⁴

(ii) 5x² – 3x by x

= 5x – 3

(iii) 10x³y – 9xy² – 4x²y² by xy 

= 10x² – 9y – 4xy 

(iv)

3y³ – 9ay² – 6ab²y by -3y

(v)

x⁵ – 15x⁴ – 10x² by -5x²

(vi) 12a² + ax – 6x² by 3a – 2x 

= 4a + 3x 

(vii)

6x² – xy – 35y² by 2x – 5y

= 3x + 7y

(viii) x³ – 6x² + 11x – 6 by x² – 4x + 3 

= x – 2 

(ix)

m³ – 4m² + m + 6 by m² – m – 2

= m – 3

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Fundamental Concepts ICSE Class-7 Exe- 11 E

Simplify

Question 1.

Answer

Question 2.

Answer

Question 3.

Answer

Question 4.

Answer

Question 5.

Answer

Question 6.

Answer

Question 7.

Answer

(LCM of 2, 3, 5 = 30)

Question 8.

Answer

(LCM of 5, 4 = 20)

Question 9.

Answer

Question 10.

Answer

Question 11.

Answer

Question 12.

Answer

Question 13.

Answer

Question 14.

Answer

Question 15.

Answer

Question 16

Answer

Question 17.

Answer

Question 18.

Answer

.(LCM of 3,2 = 6)

Question 19.

Answer

Question 20.

Answer

Question 21

.

Answer

Question 22.

Answer

Question 23.

Answer

Question 24.

Answer

Question 25.

Answer

Question 26.

Answer

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Exe – 11 F Concise Selina Maths ICSE Class-7

Enclose the given terms in brackets as required :

Question 1.

x – y – z = x-{…….)
Answer

x – y – z = x – (y + z) 

Question 2.

x² – xy² – 2xy – y² = x² – (…….. )
Answer

x² – xy² – 2xy – y²

= x² – (xy² + 2xy + y²)

Question 3.

4a – 9 + 2b – 6 = 4a – (…….. )
Answer

4a – 9 + 2b – 6

= 4a – (9 – 2b + 6) 

Question 4.

x² -y² + z² + 3x – 2y = x² – (…….. )
Answer

x² -y² + z² + 3x – 2y

= x² – (y² – z² – 3x + 2y) 

Question 5.

– 2a² + 4ab – 6a²b² + 8ab² = – 2a (……… )
Answer

– 2a² + 4ab – 6a²b² + 8ab²

= – 2a (a – 2b + 3ab² – 4b²)

 

Simplify :

Question 6.

2x – (x + 2y- z)
Answer

2x – (x + 2y- z)

= 2x – x – 2y + z

= x – 2y + z

Question 7.

p + q – (p – q) + (2p – 3q)
Answer

p + q – (p – q) + (2p – 3q)

= p + q – p + q + 2p – 3q

= 2p – q

Question 8.

9x – (-4x + 5)
Answer

Question 9.

6a – (- 5a – 8b) + (3a + b)
Answer

6a – (- 5a – 8b) + (3a + b)

= 6a + 5a + 8b + 3a + b

= 6a + 5a + 3a + 8b + b

= 14a + 9b

 Question 10.

(p – 2q) – (3q – r)
Answer

(p – 2q) – (3q – r)

=p – 2q – 3q + r

=p – 5q + r

Question 11.

9a (2b – 3a + 7c)
Answer

9a (2b – 3a + 7c)

= 18ab – 27a² + 63ca

Question 12.

-5m (-2m + 3n – 7p)
Answer

– 5m (-2m + 3n – 7p)

= – 5m x (-2m) + (-5m) (3n) – (-5m) (7p)

= 10m² – 15mn + 35 mp.

Question 13.

-2x (x + y) + x²
Answer

– 2x (x + y) + x²

= -2x × x + (-2x)y + x²

= – 2x² – 2xy + x²

= – 2x² + x² – 2xy = – x² – 2xy

Question 14.

Answer

= b × 2b – b × (1/b) – 2b×b+2b×(1/b)

= 2b² – 1 – 2b² + 2

= 2b² – 2b² – 1 + 2

= 1

Question 15.

8 (2a + 3b – c) – 10 (a + 2b + 3c)
Answer

8 (2a + 3b – c) – 10 (a + 2b + 3c)

= 16a + 24b – 8c – 10a – 20b- 30c

= 16a – 10a + 24b – 20b – 8c – 30c

= 6a + 4b – 38c

Question 16.

Answer

= a² + 1 – b² + 1 – c² – 1

= a² – b² – c² + 1

Question 17.

5 x (2x + 3y) – 2x (x – 9y)
Answer

5 x (2x + 3y) – 2x (x – 9y)

= 10x² + 15xy – 2x² + 18xy

= 10x² – 2x² + 15xy+ 18xy

= 8x² + 33xy

Question 18.

a + (b + c – d)
Answer

a + (b + c – d)

= a + (b + c – d)

= a + b + c – d

Question 19.

5 – 8x – 6 – x
Answer

5 – 8x – 6 – x

= 5 – 6 – 8x – x

= -1 – 7x

Question 20.

2a + (6-  )
Answer

2a + (6-  )

= 2a + (b – a + b)

= 2a + b – a + b

= a + 2b

Question 21.

3x + [4x – (6x – 3)]
Answer

3x + [4x – (6x – 3)]

= 3x + [4x – 6x + 3]

= 3x + 4x – 6x + 3

= 3x + 4x – 6x + 3

= 7x – 6x + 3

= x + 3

Question 22.

5b – {6a + (8 – b – a)}
Answer

5b – {6a + (8 – b – a)}

= 5b – 6a – 8 + b + a

= -6a + a + 5b +b – 8

= -5a + 6b – 8

Question 23.

2x-[5y- (3x -y) + x]
Answer

2x – [5y – (3x – y) + x]

= 2x – {5y – 3x +y + x}

= 2x – 5y + 3x -y – x

= 2x + 3x – x – 5y – y

= 4x – 6y

Question 24.

6a – 3 (a + b – 2)
Answer

6a – 3 (a + b – 2)

= 6a – 3a – 3b + 6

= 3a – 3b + 6

Question 25.

8 [m + 2n-p – 7 (2m -n + 3p)]
Answer

Question 26.

{9 – (4p – 6q)} – {3q – (5p – 10)}
Answer

Question 27.

2 [a – 3 {a + 5 {a – 2) + 7}]
Answer

2 [a – 3 {a + 5 {a – 2) + 7}]

= 2 [a – 3 {a + 5a – 10 + 7}]

= 2 [a – 3a – 15a + 30 – 21]

= 2a – 6a – 30a + 60 – 42

= 2a – 36a + 60 – 42

= -34a + 18

Question 28.

5a – [6a – {9a – (10a –   )}]
Answer

5a – [6a – {9a – (10a –   )}]

= 5a – [6a – {9a – (10a – 4a + 3a)}]

= 5a – [6a – {9a – 10a + 4a – 3a}]

= 5a- [6a – 9a + 10a – 4a + 3a]

= 5a – 6a + 9a – 10a + 4a – 3a

= 5a + 9a + 4a – 6a – 10a – 3a

= 18a – 19a

= – a

Question 29.

9x + 5 – [4x – {3x – 2 (4x – 3)}]
Answer

9x + 5 – [4x – {3x – 2 (4x – 3)}]

= 9x + 5 – [4x – {3x – 8x + 6}]

= 9x + 5 – [4x – 3x + 8x – 6]

= 9x + 5-4x + 3x – 8x + 6

= 9x + 3x – 4x – 8x + 5 + 6

= 12x – 12x + 11

= 11

Question 30.

(x + y – z)x + (z + x – y)y – (x + y – z)z

Answer

(x + y – z)x + (z + x – y)y – (x + y – z)z

= x²+ xy – zx + yz + xy -y² – zx – yz + z²

= x² – y² + z² + 2xy – 2zx

Question 31.

-1 [a-3 {b -4 (a-b-8) + 4a} + 10]
Answer

-1 [a – 3 {b – 4 (a – b – 8) + 4a} + 10]

= -1 [a – 3 {b – 4{a – b – 8) + 4a} + 10]

= -1[a – 3 {b – 4a + Ab +32 + 4a} + 10]

= -1 [a – 3b + 12a – 126 – 96 – 12a + 10]

= -a + 3b – 12a + 12b + 96 + 12a – 10

= -a-12a + 12a+ 3b+ 12b – 96 – 10

= – a + 15b – 106

Question 32.

Answer

Question 33.

10 – {4a – (7 – ) – (5a – )}
Answer

10 – {4a – (7 – ) – (5a – )}

= 10 – {4a – (7 – a + 5) – (5a – 1 – a)}

= 10- {4a -(12 – a) -(4a – 1)}

= 10 – {4a – 12 + a- 4a + 1}

= 10 – 4a + 12 – a + 4a – 1

= 10 + 12 – 1 – 4a – a + 4a

= 21 – a

Question 34.

7a- [8a- (11a-(12a- )}]
Answer

7a- [8a- (11a-(12a- )}]

= 7a – [8a – {11a – (12a – 6a + 5a)}]

= 7a – [8a – {11a – (17a – 6a)}]

= 7a – [8a – {11a – (11a)}]

= 7a – [8a – {11a – 11a}]

= 7a – 8a

= – a

Question 35.

Answer

= 8x – [4y – {4x + (2x – 2y + 2x)}]

= 8x – [4y – {4x + (4x – 2y)}]

= 8x – [4y – {4x + 4x – 2y}]

= 8x – [4y – {4x – 4x + 2y}]

= 8x – [- 8x + 6y]

= 8x + 8x – 6y

= 16x – 6y

Question 36.

x-(3y-  +2z- )
Answer

x-(3y-  +2z- )

= x – (3y – 4z + 3x + 2z -5y + 7x)

= x – (- 2y – 2z + 10x)

= x + 2y + 2z – 10x

= – 9x + 2y + 2z

— End of Fundamental Concepts Solutions :–

Return to – Concise Selina Maths Solutions for ICSE Class -7 

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